The temperature of water in the Gulf of Mexico can rise by about 20 $\circ$ F with the onset of the summer.However,the most significant temperature rise is confined to the top 2-3 meters of water.Assuming that it takes 45 days for the temperature to increase by 20 $\circ$ F,what is the average energy absorbed per day by water in a volume measuring 1 m $\times$ 1 m $\times$ 2.5 m?

Question

Quizplus · Accepted Answer

The amount of heat energy absorbed by the water can be calculated using the formula:
Q = mcΔT
where Q is the heat energy, m is the mass of water in kg, c is the specific heat capacity of water (4.18 J/gC), and ΔT is the change in temperature in Celsius.

Since we know the increase in temperature and the volume of water, we can calculate the mass of water as follows:
V = l × w × h = 1 × 1 × 2.5 = 2.5 m^3
m = ρV = 1000 kg/m^3 × 2.5 m^3 = 2500 kg

Next, we can convert the temperature increase from Fahrenheit to Celsius:
20°F = 11.11°C

Finally, we can use the given time period to calculate the average energy absorbed per day:
t = 45 days = 3,888,000 seconds
Q = mcΔT = (2500 kg) × (4.18 J/gC) × (11.11°C) = 116,458,500 J
Average energy absorbed per day = Q/t = 116,458,500 J / 3,888,000 s ≈ 29.97 J/s
Average energy absorbed per day in 1 m x 1 m x 2.5 m volume = 29.97 J/s x 1 m x 1 m x 2.5 m = 74.92 J
Rounding off to two significant figures, we get the answer as 2.58x10^16 J or choice B.

The Temperature of Water in the Gulf of Mexico Can ∘\circ∘

The Temperature of Water in the Gulf of Mexico Can $\circ$